Notes on the Weil conjectures. Kevin Buzzard April 26, 2012 Written Feb 2010. 1 The definition. The idea is that the Riemann zeta function can be thought of as Y ζ(s) = (1 − #k(c)−s)−1 c where the product is over the closed points c of Spec(Z), and k(c) is the residue field of c. The analogue is this: if V is a scheme of finite type over a finite field k of size q, then we can define Y −s −1 ζV (s) = (1 − #k(c) ) c2jV j where jV j denotes the closed points of the scheme V . 1.1 Where does this converge? Well, here's a fundamental but trivial link between closed points and kn-points, with kn the extension of k of degree n: a closed point c with k(c) of size qd (let's call such c a \degree d point") contributes 0 to V (kn) if d - n (because there are no field maps kd ! kn) and d points if d j n (because in this case there are d field maps). Hence if V is a union of M affine opens, each of which is a closed subspace of some big affine N-space, then a hugely crude bound for the number Nd N of degree d closed points is Mq , because this number is M times the size of A (kd). So now we use the old trick: the product of (1+xn) converges (to something non-zero) iff σnxn converges, so if P −ds Cd denotes the number of closed points of degree d then we need to check d≥1 Cdq converges, −ds −ds log(q) −dσ log(q) −dσ Nd and jq j = je j = e = q , where σ = Re(s). So, because Cd ≤ Mq , we have P Nd −dσ N−σ absolute convergence if d≥1 q q converges, so in particular if jq j < 1, that is Re(s) > N, then the power series converges. 1.2 A more algebraic way of looking at ζV (s). If Cd is the number of points of degree d, then Y −ds −Cd ζV (s) = (1 − q ) d≥1 and if u = q−s then this is Y C1(C1 + 1) Z (u) := (1 − ud)−Cd = 1 + C u + + C u2 + ::: V 1 2 2 d≥1 which is clearly in Z[[u]] (and, as we've just seen, converges for juj sufficiently small). Moreover, for u very small, we have ZV (u) ' 1 + C1u = 1 + N1u, where N1 is the size of V (k). 1 1.3 How to relate ZV (u) to number of kn-points. This is just algebra. Here's the trick. If Nn is #V (kn) then, because a closed point of degree d contributes d points to Nn if d j n (and none otherwise) we have X Nn = dCd: djn Hence (with 0 denoting derivative with respect to u) 0 uZV (u) 0 = u(log(Zv(u))) Zv(u) X d 0 = u ((−Cd log(1 − u ))) d≥1 X −dud−1 = u −C d 1 − ud d≥1 X d d −1 = Cddu (1 − u ) d≥1 X md = Cddu d;m≥1 X X n = ( Cdd)u n≥1 djn X n = Nnu n≥1 So the generating function for the number of points (which I can sometimes work out, e.g, for projective space or a CM elliptic curve) is related to ZV (u). P n 2 How to go the other way? Say Gv(u) = n≥1 Nnu = N1u + N2u + ::: is the generating R function. Then (GV (u)=u)du (the formal integral) is log(ZV (u)), which has no constant term, so if we define the integral to have no constant term then Z G (u) Z (u) = exp V du V u 1.4 How does changing the base field change things? 0 0 0 0 n If k =k is a finite extension of degree n, and ZV 0 (u ) is the resulting function, with u = u , then 0 0 there's a simple relation between ZV 0 (u ) and ZV (u). Indeed, a closed point of degree d in V gives 0 0 rise to a kd ⊗k k -valued point of V and hence, if e = gcd(d; n), e points each of degree dn=e. Now d −1 dn=e −e an easy calculation shows that the factor (1 − u ) in ZV (u) turns into (1 − u ) which is Q d −1 just ζ (1 − (ζu) ) , with ζ running through the nth roots of unity. We conclude 0 n Y ZV 0 (u ) = ZV (ζu): ζn=1 2 Elementary Examples. 2.1 Projective spaces. We can now compute our first zeta functions. For projective 0-space let's do it the hard way, for a P n sanity check. We have GV (u) = n≥1 u = u=(1−u), and the integral of 1=(1−u) is − log(1−u), −1 so ZV (u) = (1 − u) , which of course can easily be seen from the definition (this is a case where I can compute the zeta function from its definition!) and I get a check on the formula above. 2 P n n Now let's try projective 1-space. We have GV (u) = n≥1(1 + q )u which is u=(1 − u) + qu=(1 − qu). Divide by u to get 1=(1 − u) + q=(1 − qu). Integrate to get − log(1 − u) − log(1 − qu). −1 −1 Exponentiate and get ZV (u) = (1 − u) (1 − qu) . This to me seems a bit less obvious. It says, for example, Y (1 − udeg(f)) = 1=(1 − qu) f2k[X] monic irreducible as a power series in u (or a complex function of a sufficiently small u 2 C). This idea generalises to projective r-space: it gives GV (u) = u=(1 − u) + qu=(1 − qu) + ::: + r r q u=(1 − q u), and dividing by u and integrating and exponentiating gives, I guess, ZV (u) = (1 − u)−1(1 − qu)−1 ::: (1 − qru)−1. That's pretty awesome. Now here's a second way of computing the zeta function of projective 1-space, which we'll use to great effect for more general curves later on. Let V be projective 1-space. We have Q d(c) −1 ZV (u) = c(1 − u ) , the product over the closed points, so −1 X d(D) ZV (u) = (1 − u ) u D≥0 where the sum is now over all the positive divisors on A1. But on A1 every divisor is principal with a unique monic generator, so −1 X m m ZV (u) = (1 − u) q u = 1=(1 − u)(1 − qu): m≥0 We'll use a generalisation of this trick when dealing with more general curves below. While I'm here let me note that if V is projective 1-space then uZV (u) is invariant under u 7! 1=(qu), or, −s equivalently, that ζV (s)q is invariant under s 7! 1 − s. 2.2 Hyperelliptic curves. Now we move onto some less trivial examples. Let's take an elliptic curve E : y2 = f(x) with f monic in k[x] and of degree 3, and with no repeated root, and p 6= 2. Here is how Artin dealt with this case (and more generally with the hyperelliptic case when p 6= 2). At infinity we have one point (note: if I were dealing with f of smaller degree I might have two points) of degree 1. All the other points correspond to some prime ideal of the ring R := k[x; y]=(y2 − f(x)). Now if P is a non-zero prime ideal of k[x], then P factors in R into prime ideals. We know that P remains prime if f is not a square mod P , that P is ramified if f 2 P , and P splits as P = Q1Q2 if f is a square mod P . Now if P = (g) with g monic in k[x], then quadratic reciprocity (which Artin has proved in his thesis) says, amongst other things, that if q = 1 mod 4 then f is a square mod g iff g is a square mod f! And if q = 3 mod 4 then there are easily-controlled sign issues. Anyway let's not get into this. Let χ be the function on prime ideals such that χ(P ) = 0 if f 2 P , χ(P ) = 1 if f is a square mod P but not in P , and χ(P ) = −1 if f is not a square mod P . Then −1 Y d(P ) −1 d(P ) −1 ZE(u) = (1 − u) (1 − u ) (1 − χ(P )u ) P ⊂k[x] where d(P ) is the degree of P , and the product is over all prime ideals of k[x]. We can compute the product of the first terms by the Weil conjectures for P1 (it's 1=(1 − qu)), and if we do it in the 2nd way I explained (by expanding the product as a sum) we get a clue as to how to proceed with the other product. Indeed, same technique shows that the other product is P d(M) P m P M χ(M)u = m≥0 σmu , with σm = M:d(M)=m χ(M). But now here's the bombshell: σm = 0 if m ≥ 3! For by quadratic reciprocity χ(M) only depends on g mod f, where M = (g) (at least if q = 1 mod 4; if it's 3 mod 4 then it also depends on the parity of the degree of g but this can be dealt with) and the sum over all non-zero residues of the character is 0.1 1 1I never checked this carefully.